a. \(\sqrt{\frac{3}{7}}=\sqrt{\frac{3\cdot7}{7^2}}=\frac{\sqrt{21}}{7}\)

b.\(\sqrt{\frac{7}{20}}=\sqrt{\frac{7\cdot5}{4\cdot5\cdot5}}=\frac{\sqrt{35}}{2\cdot5}=\frac{\sqrt{35}}{10}\)

c.\(\sqrt{\frac{11}{12}}=\sqrt{\frac{11\cdot3}{4\cdot3\cdot3}}=\frac{\sqrt{33}}{2\cdot3}=\frac{\sqrt{33}}{6}\)

d.\(\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{3}}=\sqrt{\frac{3\left(\sqrt{3}-\sqrt{2}\right)^2}{3\cdot3}}=\frac{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{3}=\frac{3-2\sqrt{3}}{3}\)

a. ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(B=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

b.

\(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}=0\\ \Leftrightarrow\frac{2\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}+1\right)}=0\\ \Leftrightarrow\frac{2\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}=0\\ \Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}=0\\ \Leftrightarrow\sqrt{x}-3=0\\ \Leftrightarrow x=9\left(tm\right)\)

c.

\(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\\ \Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}< 0\\ \Leftrightarrow\sqrt{x}-3< 0\\ \Leftrightarrow x< 9\)

Vậy với \(0\le x< 9;x\ne1\)thì B < 1/2

ĐKXĐ: \(x\ge4\)

\(x-4\sqrt{x-4}=\left(\sqrt{x-4}-2\right)^2\\ \Leftrightarrow x-4\sqrt{x-4}=\left(x-4\right)-2\cdot\sqrt{x-4}\cdot2+4\\ \Leftrightarrow x-4\sqrt{x-4}-x+4+4\sqrt{x-4}-4=0\)

\(\Leftrightarrow0x=0\) (luôn đúng với mọi giá trị của x)

Vậy với mọi \(x\ge4\) thì ......

a)

\(A=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\\ =\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\\ =\frac{\sqrt{a}-1}{\sqrt{a}}\)

b) Ta có: \(A=\frac{\sqrt{a}-1}{\sqrt{a}}=\frac{\sqrt{a}}{\sqrt{a}}-\frac{1}{\sqrt{a}}=1-\frac{1}{\sqrt{a}}\)

Với mọi a>0 và a≠1 ta có \(\sqrt{a}>0\Leftrightarrow\frac{1}{\sqrt{a}}>0\)

\(\Rightarrow A=1-\frac{1}{\sqrt{a}}< 1\left(đpcm\right)\)

c)

\(A=1-\frac{1}{\sqrt{a}}=\frac{1}{2}\Leftrightarrow\frac{1}{\sqrt{a}}=\frac{1}{2}\Leftrightarrow\sqrt{a}=2\Leftrightarrow a=4\left(tm\right)\)

Vậy.......

\(C=5x^2-7x+4\\ =5\left(x^2-\frac{7}{5}x\right)+4\\ =5\left(x^2-2\cdot x\cdot\frac{7}{10}+\left(\frac{7}{10}\right)^2\right)+\frac{31}{20}\\ =\left(x-\frac{7}{10}\right)^2+\frac{31}{10}\ge\frac{31}{10}\forall x\)

Vậy Min C = \(\frac{31}{10}\)khi \(x=\frac{7}{10}\)

\(D=x^2+y^2-2x-4y-6\\ =\left(x^2-2x+1\right)+\left(y^2-4y+4\right)-11\\ =\left(x-1\right)^2+\left(y-2\right)^2-11\)

Ta thấy \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(y-2\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow D=\left(x-1\right)^2+\left(y-2\right)^2-11\ge-11\forall x,y\)

Vậy min D = -11 khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

1.Check your work on the screen. Onc you're satisfied, you can back up copy.

2.If the printer runs out of paper, a warning light indicates that you need some more.

3.The boss said it was nobody's fault but we put it down to bad programming on the part of those "experts" at head office.

4.The program took up a lot of dish space so there wasn't much room for anything else.

5.Calculations which used to take ages can now be worked out in a few seconds.

nối hai câu đơn đã cho thành 1 câu ghép,sử dụng liên từ thích hợp

1)Jim was absent from class yesterday.he was suffering from a bad cold

=> Jim was absent from class yesterday because he was suffering from a bad cold.

2)They decided to move their house.they don't have to travel a long distance to work

=> They decided to move their house so (so that) they don't have to travel a long distance to work

3)mary hurt her fingers yesterday.she was preparing dinner

=>Mary hurt her fingers yesterday while(when) she was preparing dinner

4)At 4pm yesterday,my mother was cleaning the house.I was looking after my younger brother

=> At 4pm yesterday, my mother was cleaning the house while I was looking after my younger brother.

5)They decided to go an excursion.They want to get away from work stress

=> They decided to go an excursion since/because they want to get away from work stress.

6)my father taught me how to use the computer.before that, I did't know how to use it

=> I didn't know how to use the computer until my father taught me how to use it

7)I arrive at the station.I will call you right after that

=> I will call you as soon as I arrive at the station.

8)My brother is very out-going.I am quite reserved

=> My brother is very out-going but I am quite reserved.

9)Mr.peter was walking home.he was robbed of his wallet

=> Mr.peter was walking home when he was robbed of his wallet.

10)I don't have a pet.my mother doen't allow me to have one

=> I don't have a pet because my mother doen't allow me to have one.

Cái này áp dụng công thức là ra thôi

Ta có \(\tan a\cdot\cot a=1\Leftrightarrow\frac{\sqrt{2}}{2}\cdot\cot a=1\Leftrightarrow\cot a=1:\frac{\sqrt{2}}{2}=\sqrt{2}\)

Có \(\cot a=\frac{\cos a}{\sin a}\Leftrightarrow\sqrt{2}=\frac{\cos a}{\sin a}\Leftrightarrow\cos a=\sqrt{2}\sin a\)

Lại có:

\(\sin^2a+\cos^2a=1\Leftrightarrow\sin^2a+\left(\sqrt{2}\sin\right)^2=1\\ \Leftrightarrow\sin^2a+2\sin^2a=1\\ \Leftrightarrow3\sin^2a=1\\ \Leftrightarrow\sin^2a=\frac{1}{3}\Leftrightarrow\sin a=\frac{1}{\sqrt{3}}\)

Từ đó suy ra \(\cos a=\sqrt{2}\sin a=\sqrt{2}\cdot\frac{1}{\sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}}=\sqrt{\frac{2}{3}}\)